Discharge Temp on a Gas Jack 1

[zdas04]

I was looking at a Gas Jack compressor from Compressco (it is an integral machine based on a Ford V-8 engine, one bank of pistons is used to drive the unit, the other bank has had head modifications and is a compressor) and I got confused.

The conditions were:
Suction 4 psig
Discharge 175 psig
Atmospheric pressure 12 psia
Volume flow rate 150 MSCF/d
Suction Temp 105F
Discharge Temp 310F
k=1.28

When I do the isentropic temp arithmetic I get that this operation is almost 12 ratios (the manufacturer claims that the unit is good for 20 ratios in one stage, and it looks like they may be right) and the discharge temp should be 507F.

The valve arrangement has both the suction and discharge valves in the same plate so the inlet stream and the outlet stream act like a gas-to-gas counterflow heat exchanger.

Also the jacket-water cooling that was provided by Ford is still in place.

It looks to me like with the 11.7^(0.28/1.28) = 1.712, any temperature rise from the gas-to-gas heat exchanger would make the discharge temp even higher.

Bottom line is that the preponderance of the lost heat must either through the jacket water cooling (in this case I've transferred almost 200F out of the process) or the isentropic math means nothing in this case.

My question is: does an isentropic model mean anything at all with a temperature loss of this magnitude? If not, what sort of model might better predict temperature out of the cylinders?

David

 

[dcasto]

It wouldm only take .6 gpm of radiator water to cool the gas down from 500 F to 300 F on 100 mscfd.

The isentropic equation is used as a starting point for discharge temperature, but a PD compressor is only 75 to 85% efficient (mainly valve and other gas flow losses make the effiency drop). All the old compressors used water cooled jackets to help the compressor remove heat from the added discharge temperature due to the efficiency. So, the gas temperature from the old compressors matched the isentropic compression temperature quite well.

Along comes Ariel, and no more water cooled cylinders and some heat exchange. Now run your iso equation against what ariel predicts. Read the GPSA data book and the authors point this out and recommend switching to a polytropic equation to predict discharge temperature. In the late 70's I picked up some equation from a paper presented on estimating the actual discharge temperature from non water cooled cylinders her it is (as cut from the FOTRAN program)

VN is k TS and TD are suction and discharge temps and R is apparent ratio with valve losses.

1030 C1P10 = 1.10675
1040 CON = (VN(L) - 1!) / VN(L)
1530 IF WC = 0 THEN 1560
1540 TD(LST) = (TS(LST) + 460!) * R ^ CON - 460!
1550 GOTO 1570
1560 TD(LST) = (TS(LST) + 460!) * ((R ^ CON - 1!) * C1P10 * (1! + .078 / (R - .793)) + 1!) - 460

Back in the 70's some operators cut the flow through water cooled cylinders and just kept them full of EG/Water mix. The only thing that happen was the discharge temperature approach the equation in line 1560 above.

As for the gasjac, I've seen a lot of them (starting in 1998 or 99) and never dug into their operation. If the unit uses the cam and valves on the engine, they could approach 90% effiency.

 

[pmover]

dcasto,

a slight correction to program language - it is gwbasic/basica and not fortran.

otherwise, thanks!
-pmover

 

[dcasto]

you are right, the original was basic on my VIC 20 and I converted it to FORTRAN for a PC when I copywrited it in 82.

 

[zdas04]

dcasto,
Thanks, I can't think about this today but I'll look at it over the weekend, I know I'll have some questions then (its killing me not to be able to mess with it now, but I have other deadlines).

David

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[dcasto]

zdas04, by the way, we reside just a few miles apart.

 

[zdas04]

I thought you were in Northen Colorado. Have you moved to the San Juan? Send me an e-mail at the Contact Me address on my web page maybe we can get together for a beer.

David

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[zdas04]

dcasto,
Well, I got a chance to look at the equation and I'm more confused than ever.

The C1P10*(1+.078/(ratio-.793)) term evaluates to 1.115 in this scenario, when you multiply the isentropic exponent minus 1 times 112% and then add the 1 back you get a discharge temp that is 5% higher (in absolute terms) than the isentropic equation. I observed a temperature that is 27% lower than isentropic.

For 4 ratios it is 3% higher than isentropic.

Playing with numbers, if I reduce C1P10 from 1.10675 to 0.506 then I get the 310F discharge temp that I observed, but I don't know the physical significance of this term.

Reducing the .078 term or increasing the .793 term makes the value in the parenthesis approach 1.0. With C1P10 greater than 1.0 this can't result in reducing the discharge temp.

Do you remember the origin of this equation? I'd sure like to see what I'm missing here.

David

 

[dcasto]

I don't remember, look at the calculation for HP:

HP =(50.8*(R ^ CON - 1!)*(R - .715))/(CON*(R - .793))*ZAVG

Some of the terms are similar. Looking at the GPSA databook's equations, the terms with (R - value) are similar to the assumed efficency terms that you use in those published equations. I don't remember where the added terms come from, but they kind of match the table to convert to polytropic from isentropic effiency and you'd expect about 3% change.